ADGEOM 209

Dear friends:

 

According to description of X(165) = centroid of the excentral triangle in ETC, if A’B’C’ is the pedal triangle of X(165) then |AB’| + |AC’| = |BA’| + |BC’| = |CA’| + |CB’|.

 

As a matter of fact, there are other three points satisfying this condition, although none of them are centers of a triangle.

These points are:

A* =  f(a,b,c) : g(a,b,c) : h(a,b,c)

B* =  h(b,c,a) : f(b,c,a) : g(b,c,a)

C* =  g(c,a,b) : h(c,a,b) : f(c,a,b)

where (trilinears)

f(a, b, c) =  (b-c)^2 + a*(a-2*b-2*c)

g(a, b, c) = (a-b)^2 + c*(-3*c+2*a+2*b)

h(a, b, c) = (a-c)^2 + b*(-3*b+2*a+2*c) = g(a, c, b)

 

Q: Extraversion? I can’t find a precise definition of what is it.  May someone please enlighten me about this?

 

Notes:

1)       A*B*C* and ABC are perspective. The perspector is X(3062) [isogonal conjugate of X(165)]

2)       A*B*C* and the excentral triangle of ABC are perspective with perspector X(165).

3)       The line (B*C*) passes through the A-excenter, (C*A*) passes through de B-excenter and (A*B*) passes through the C-excenter. Therefore, A*B*C* is the X(165)-anticevian triangle of the excentral triangle of ABC.

4)       The circumcircle of A*B*C* has center I (incenter of ABC) and radius 4*R (R=circumradius of ABC). So, it’s easy to find A*, B* and C*: A* is the intersection of this circle and the semi-line [X(3062)-A) and cyclically for B* and C*

5)       The exsimilcenter of  the circumcircles of A*B*C* and ABC is X(165)

6)       The insimilcenter I’’ of  the circumcircles of A*B*C* and ABC has trilinears: 5 cos(A)+cos(B)+cos(C) – 1 : :

I’’ = Midpoint of (X(3522), X(3616))  [X(3522) = EULER LINE INTERCEPT OF LINE X(8)X(165)]

I’’ = Reflection of X(1698) on X(631) [X(1698) = INSIMILICENTER( BEVAN CIRCLE , NINE-POINT CIRCLE )]

I” lies on lines (X(I), X(J)) for these [I,J]: [1, 3], [2, 4297], [4, 3624], [9, 3207], [10, 3523], [20, 1125], [21, 3062], [41, 572], [63, 5303], [78, 5223], [100, 4853], [104, 4866], [200, 2975], [214, 1768], [355, 549], [376, 946], [405, 1750], [499, 3586], [515, 631], [516, 3522], [551, 962], [573, 1475], [581, 5313], [936, 993], [944, 3524], [952, 4668], [956, 4882], [958, 5438], [991, 1193], [995, 4300], [997, 5267], [1001, 2951], [1006, 1490], [1012, 5259], [1030, 3554], [1055, 3731], [1210, 4305], [1483, 3654], [1572, 5206], [2136, 4421], [2801, 3876], [3085, 4311], [3086, 4304], [3146, 3817], [3430, 5429], [3485, 4312], [3486, 3911], [3487, 4355], [3553, 5124], [3622, 4301], [3636, 5493], [3651, 5426], [3652, 5428], [3655, 4677], [3751, 5085], [4189, 4512], [4220, 5272], [4293, 5290], [4308, 5281], [4313, 5265], [4511, 4652]

7)       The NPC of A*B*C* = X(40) of ABC [Bevan point]

8)       The NP circles of A*B*C* and ABC have insimilcenter X(165) and exsimilcenter X(1)

9)       The incenter of A*B*C* is X(167)=Nagel point of excentral triangle.

10)   The Euler lines of ABC and A*B*C* meet at O [circumcenter of ABC]=X(140) of A*B*C* [midpoint of (O,NPC) of A*B*C*]

11)   If A’B’C’ is the pedal triangle of A*  then the required distances are |AB’| + |AC’| = |BA’| + |BC’| = |CA’| + |CB’| = 8*R*cos(A/2)*cos(B/2)*cos(C/2) and equally for the pedal triangles of B* and C*.

 

Regards

César Lozada